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Pavel [41]
2 years ago
9

A city uses a water tower to store water for times of high demand. When demand is light, water is pumped into the tower. When de

mand is heavy, water can flow from the tower with- out overwhelming the pumps. To provide water at a typical 350 kPa gauge pressure, how tall must the tower be?
Physics
1 answer:
love history [14]2 years ago
6 0

Answer:

The height of the tower will be 35.714 m

Explanation:

We have given gauge pressure P=350kPa=350\times 10^3Pa

Density of water \rho =1000kg/m^3

We have to find the height of the tower h

We know that gauge pressure is given by P=\rho gh

350\times 10^3=100 0\times 9.8\times h

h=35.714m

So the height of the tower will be 35.714 m

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Cathode ray tubes in old television sets worked by accelerating electrons and then deflecting them with magnetic fields onto a p
Roman55 [17]

Answer:

B = 0.046T

Explanation:

given

size of the screen = 51.2cm

distance from center = 11.1cm

region of magnetic field = 1.00cm

V= 22000V= 22kV

 

3 0
2 years ago
A 36,287 kg truck has a momentum of 907,175 kg • m/s. What is the trucks velocity.
Airida [17]

Momentum = Mass x Velocity

Put the values where they belong and solve for Velocity.

In this case, since Mass is being multiplied by Velocity, to solve for be Velocity you would divide both sides by Velocity.  The velocity will equal the momentum divided by the mass.

4 0
2 years ago
While ice skating, you unintentionally crash into a person. Your mass is 60 kg, and you are traveling east at 8.0 m/s with respe
kaheart [24]

Answer:

6.18 m/s

Explanation:

Roller skate collision

The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;

x-axis component form (+x east);

P_{Miy} + p_{Piy} + j_{y}= P_{Mfy} +P_{pfy}

m_{Mu_{Miy}+ m_{pu_{piy}}+0=(m_{M}+m_{p})V_{f} sinФ

60 ·8 + 0 = (60 + 80)V_{f}sinФ

480 = 140V_{f} sinФ................. (I)

y-axis component form (+y north);

P_{Mix} + p_{Pix} + j_{x} = P_{Mfx}+ P_{pfx}

m_{Mu_{Mix}+ m_{pu_{pix}}+0=(m_{M}+m_{p})V_{f} cosФ

0 + 80.9 = (60 + 80)V_{f}cosФ

 720= 140V_{f}cosФ

140Vf=\frac{720}{cos}Ф......................................(2)

 Substituting (2) into (1) to give the angle;

 480 = 720tan Ф

Ф = arctan(0.67) =33.69°.......................(3)

Evaluating (1) with (3) gives the velocity magnitude

480 = 140Vfsin 33.69°

Vf=6.18 m/s

note 1:

This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.

 

7 0
2 years ago
a box weighing 155 N is pushed horizontally down the hall at constant velocity. the applied force is 83 n what is the coefficien
meriva

Answer:

μ = 0.535

Explanation:

On a level floor, normal force = weight.

N = W

Friction force = normal force × coefficient of friction.

F = Nμ

Substitute:

F = Wμ

83 = 155μ

μ = 0.535

Round as needed.

8 0
2 years ago
Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov
frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v_{1} be the velocity of ball A along positive x axis

v_{2} be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v_{1}

∴  v_{1} = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v_{2}

2 = u +  v_{2} → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 =  v_{2} - u → equation 2

By adding both equations 1 and 2 we get

v_{2} = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

                         = 8·85×m J

Total momentum = m×√(2² + 3·7²)

                             = m× 4·21 kgm/s

3 0
2 years ago
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